Mechanics help needed

[arlery]

Q. A particle is projected upwards with a speed of 14m/s. Find for how long it's above 2m.

 

[ahmed t]

n

 

[arlery]

Q. A ball is thrown vertically upward with a speed of 29 m/s. It hits the ground 6 seconds later. By modeling the ball as a particle find the height above the ground from which it was thrown.

 

[ahmed t]

2 seconds

 

[arlery]

What about:

Q. A ball is thrown vertically upward with a speed of 29 m/s. It hits the ground 6 seconds later. By modeling the ball as a particle find the height above the ground from which it was thrown.

 

[ahmed t]

first find the hieght for which the ball reaches it highest limit
using V^2=U^2 =2as
and make V=0
u shud find the distance to be 42.05
insert this in any equation and find the time taken which shub be 2.9
to find the distanc it then falls till it hits the ground we use a time of 6-2.9=3.1
insert this in an equation and find out the distance
it shud be48.05
hence distance above ground the particle was thrown is 6m
hope it helps and hope im right

 

[ahmed t]

wat was the anser for the first question u asked

 

[arlery]

2.5 s

 

[ahmed t]

ye thnx i got that one correct
did u get my explanation??
sorry but im hunting for the nabeel elias book cant seem to find it!!!!

 

[ahmed t]

have u noticed that we both got ranked up to elite lately

 

[arlery]

Yeah, once you reach 200 posts you're ranked an XPF Elite :Yahoo!:
I hope you find your Nabeel Elias book !!

 

[arlery]

Yeah they're right !!! =) and yeah you explain really well. :Bravo:

 

[ahmed t]

found it post ur doubts here!

 

[arlery]

YAYYY !!!! :Yahoo!:

Kinematics. (Exercise 1),
Q 13 c, Q 15, 14(i), 16, 23 c

 

[ahmed t]

for 13c its zero since the graph is horizontal line there is no movement hence no velocity

15) wen t is 5 it is in the AB straight line segment. ie the speed is the gradient of this straight line u can find out the gradien using the points of A and B
it should be 0.178

 

[arlery]

BTW how was your P12?

 

[ahmed t]

t=15 it is in the CD segment and as u see is a horizontal line so Zero

t=25 it is in the EF segment so all we do is find out the gradient and that will be -0.15


the tV graph
from T=0 to T=3 ther is a straight line with a positve gradient then from 3 to 11 it is a straght line then from 11 to 13 it drops to zero velocity an stays ther until T=20
then it will go under the Xaxis (negative) until T=22 and levels of in a straight line while bieng under the Xaxis
at T=34 it drops verticaly to zero

 

[ahmed t]

gd made a few silly mistakes in simeltaneous equations and one tiny mistake in the last question alrest was fine

 

[ahmed t]

14i)
from about 10 to 25 the straight line segment cause it has the highest gradient

 

[arlery]

But for 14 (i) answer is 20s